1. Cell Process 1. Cell and transport phenomena (2) Liang Yu Department of Biological Systems Engineering Washington State University 02. 12. 2013

2. Main topics Cell and transport phenomena High performance computation Metabolic reactions and C-13 validation Enzyme and molecular simulation

3. Physiological Fluid Mechanics Fluid A material that deforms continuously when subjected to a force applied tangentially to a surface Biological fluid Quite complex, exhibiting solid and liquid-like behavior and deforming in a time-dependent fashion Fluid mechanics The study of the motion of fluids in response to the application of stresses Physiological Fluid Mechanics Concerned with fluid flow within organisms and the relation between fluid flow and physiological processes

4. Conservation relations and momentum balances Fluid kinematics: describe fluid motion in time and space Eulerian description: observing motion from a fixed reference frame Lagrangian description: observing motion from a moving reference frame

5. Deference between Eulerian and Lagrangian description River bank If record the motion from the riverbank, Eulerian description (fixed reference frame ) If you are on the boat as it drifts with the current, Lagrangian description (moving)

6. Eulerian approach Describes the flow field (velocity, acceleration, pressure, temperature, etc.) as functions of position and time Do not follow particles, i.e. different particles will flow through the same point at different times Relatively low computation, most fluid mechanics textbooks and papers use this system The time-change of the velocity in such a measurement Called the partial derivative of the temperature with respect to time, or local accelerate Velocity field:

7. (x,y,z) Fluid element and properties The behavior of the fluid is described in terms of macroscopic properties: –Velocity u. –Pressure p. –Density  –Temperature T. –Energy E. Properties are averages of a sufficiently large number of molecules. A fluid element can be thought of as the smallest volume for which the continuum assumption is valid. x y z yy xx Faces are labeled North, East, West, South, Top and Bottom zz

8. Lagrangian approach Follow individual particles, positions of specified particles are the objectives Track (or follow) it as it moves, and monitor change in its properties The properties may be velocity, temperature, density, mass, or concentration, etc in the flow field The recorded properties are associated with the same fluid particle, but at different locations and at different times The time-change of the velocity in such a measurement Called material derivative or substantial derivative. It reflects time change in the velocity (or any other properties) of the tagged fluid particles as observed by an observer moving with the fluid. Lagrangian approach is also called “particle based approach”.

9. Lagrangian approach Relatively high computation, difficult to use for practical flow analysis Fluids are composed of billions of molecules Interaction between molecules hard to describe/model The above equations determine the path of the particle A if the position is specified at some initial instant in its path history

10. Acceleration Fluid velocity is a function of the three spatial coordinates x, y, z, and if the flow is unsteady, time t The total differential is defined as Acceleration is a vector defined as the rate of change of velocity Local acceleration Convective acceleration

11. Rate of change for a fluid particle Terminology: fluid element is a volume stationary in space, and fluid particle is a volume of fluid moving with the flow. A moving fluid particle experiences two rates of changes: –Change due to changes in the fluid as a function of time. –Change due to the fact that it moves to a different location in the fluid with different conditions. The sum of these two rates of changes for a property per unit mass  is called the total or substantive derivative D  /Dt: With dx/dt=u, dy/dt=v, dz/dt=w, this results in:

12. Rate of change for a stationary fluid element In most cases we are interested in the changes of a flow property for a fluid element, or fluid volume, that is stationary in space. However, some equations are easier derived for fluid particles. For a moving fluid particle, the total derivative per unit volume of this property  is given by: For a fluid element, for an arbitrary conserved property  : Continuity equationArbitrary property (for moving fluid particle) (for given location in space)

13. zero because of continuity Rate of increase of  of fluid element Net rate of flow of  out of fluid element Rate of increase of  for a fluid particle = Fluid particle and fluid element We can derive the relationship between the equations for a fluid particle (Lagrangian) and a fluid element (Eulerian) as follows:

14. Control volumes  A convenient region of space  For examining the flow of mass, momentum, and energy  Control volume is arbitrary  Control volume can be fixed in size or may change size with time  Simple geometry is square (2D) or cube (3D)

15. Continuity equation Rate of increase of mass in fluid element equals the net rate of flow of mass into element. Rate of increase is: The inflows (positive) and outflows (negative) are shown here: x y z

16. Continuity equation Summing all terms in the previous slide and dividing by the volume  x  y  z results in: In vector notation: For incompressible fluids  /  t = 0, and the equation becomes: div u = 0. Alternative ways to write this: and Change in density Net flow of mass across boundaries Convective term

17. Navier–Stokes equation (Momentum equation ) Describe the motion of fluid substances Applying Newton's second law to fluid motion Assumption the fluid stress is the sum of a diffusing viscous term (proportional to the gradient of velocity) and a pressure term - hence describing viscous flow Constant density, ρ Constant viscosity, μ Continuity (incompressible flow) A solution of the Navier–Stokes equations is called a velocity field or flow field, which is a description of the velocity of the fluid at a given point in space and time

18. Momentum equation in three dimensions Derive conservation equations for momentum for fluid particles. Then use the above relationships to transform those to an Eulerian frame (for fluid elements). Start with deriving the momentum equations from Newton’s second law: rate of change of momentum equals sum of forces. Rate of increase of x-, y-, and z-momentum: Forces on fluid particles are: –Surface forces such as pressure and viscous forces. –Body forces, which act on a volume, such as gravity, centrifugal, Coriolis, and electromagnetic forces.

19. Viscous stresses Stresses are forces per area. Unit is N/m 2 or Pa. Viscous stresses denoted by  Suffix notation  ij is used to indicate direction. Nine stress components. –  xx,  yy,  zz are normal stresses. E.g.  zz is the stress in the z- direction on a z-plane. –Other stresses are shear stresses. E.g.  zy is the stress in the y-direction on a z-plane. Forces aligned with the direction of a coordinate axis are positive. Opposite direction is negative.

20. Forces in the x-direction x z y Net force in the x-direction is the sum of all the force components in that direction.

21. Momentum equation Set the rate of change of x-momentum for a fluid particle Du/Dt equal to: –the sum of the forces due to surface stresses shown in the previous slide, plus –the body forces. These are usually lumped together into a source term S M : –p is a compressive stress and  xx is a tensile stress. Similarly for y- and z-momentum:

22. Application of Momentum balances Pressure-driven flow through a narrow rectangular channel Pressure-driven flow through a Cylindrical tube  Flow is induced by a pressure gradient, which could be produced by a pump or gravity.  These examples also examine the effect of geometry upon the flow and the shear stress.  Flow through these kinds of channels are important in many biomedical devices such as hemodialyzers and ultrafiltration.

23. Pressure-driven flow through a narrow rectangular channel Assumption: 1.The flow is pressure driven; that is, pressure varies only in the direction of flow 2.The fluid density is constant, which indicates that the fluid is incompressible 3.The flow is steady; that is, pressure, shear stress and velocity do not change with time 4.The fluid is Newtonian 5.Edge effect are neglected. To meet this assumption, we require a long, wide channel; that is, h/w<<1 and h/L <<1, where w is the width and L is the length of the channel 6.The flow is laminar

24. Pressure-driven flow through a narrow rectangular channel Parallel-plate channels are used widely to study the effect of flow on cell adhesion and cell function. Since there is no net momentum flow and the flow is steady, the sum of all forces must equal zero. The only forces arising are those due to pressure and shear stress. A momentum balance in the x direction yields

25. Pressure-driven flow through a narrow rectangular channel Dividing by ∆x∆y∆z and taking the limit as each goes to zero results in the ordinary differential equation The pressure changes only in the x direction (i.e., dp/dx=f(x)) and the shear stress changes in the y direction (i.e., dτ/dy=g(y)) Integrate f(x) to yield

26. Pressure-driven flow through a narrow rectangular channel The pressure can be specified at two locations, away from both the entrance and exit. Thus, at x=x 0, p=p 0, and at x=x L, p=p L. Defining ∆p=p 0 -p L and L=x L -x 0 to remove C 1 and C 2 Integrate τ to yield Newton’s law of viscosity

27. Pressure-driven flow through a narrow rectangular channel Substitute into Newton’s law of viscosity After integrating this equation Apply the boundary condition that u x =0 at y=±h/2

28. Pressure-driven flow through a narrow rectangular channel The velocity is a maximum at y=0 The volumetric flow rate is the integral of the velocity over the cross-sectional area

29. Pressure-driven flow through a narrow rectangular channel Analytic solution Numerical solution

30. Pressure-driven flow through a narrow rectangular channel function NewtonianFluidFlowRectangular % Pressure-driven flow through a cylindrical tube % Solve momentum equation to obtain shear stress and velocity distribution % Laminar Flow in a Horizontal rectangular channel (Newtonian Fluid) % clear all clc global deltaP L mu H deltaP = 1; L = 1.5; mu = 1.0e-3; H = 0.1; a = 0; b = H; % Sovle the problem of ODE-BVP % initialize of solution with a guess of y1(r)=0,y2(r)=0,y3(r)=0 solinit = bvpinit(linspace(a,b,100),[0 0 0]); sol = bvp4c(@ODEs,@BCfun,solinit); Use Matlab to provide analytic solution and numerical solution ∆ p= 1(Pa) H=0.1(m) L=x L -x 0 =1.5(m) µ= 1.0 x 10 -3 (Pa s, N s/m 2 )

31. Pressure-driven flow through a narrow rectangular channel % Analysis results h = sol.x/2; TauAnal = deltaP*h/L; uAnal = (deltaP*H^2/(8*mu*L))*(1-(2*h/H).^2); umAnal = deltaP*H^2/(8*mu*L); % Plot % Shear stress tau = sol.y(1,:); %tau = [0 tau]; plot(h,tau,'b-',h,TauAnal,'h-.') xlabel('Plane hight h£¬m') ylabel('Shear stress£¬kg/(m s^2)') legend('Numerical results','Analysis results') figure % Velocity distribution plot(h,sol.y(2,:),'b-',h,uAnal,'h-.') xlabel('Plane hight h£¬m') ylabel('Velocity£¬m/s') legend('Numerical results','Analysis results') % Average velocity fprintf('\tAverage velocity: um = %.4f',sol.y(3,end))

32. Pressure-driven flow through a narrow rectangular channel function dydh = ODEs(h,y) global deltaP L mu H Tau = y(1); u = y(2); um = y(3); dTaudh = deltaP/L/2; dudh = -Tau/mu/2; dumdh = u*2*h/H^2; dydh = [dTaudh; dudh; dumdh]; % ------------------------------------------------------------------ function bc = BCfun(ya,yb) bc = [ya(1); yb(2); ya(3)];

33. Pressure-driven flow through a narrow rectangular channel

34. We also can use Matlab PDEtool to provide numerical solution

35. Pressure-driven flow through a narrow rectangular channel